It is about finding kinetic energy want help in part 1 not part 2 in part 1 please tell how to find velocity?

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It is about finding kinetic energy want help in part 1 not part 2 in part 1 please tell how to find velocity?

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Answer 1 · 2018-04-28T10:13:53

(i) Driving Force of the engine $= \frac{Wattage}{velocity}$ $="Wattage"/"velocity"$
At the bottom of hill
Net force upwards $= \frac{30000}{v_{bottom}} - 1000 - m g sin θ$ $=30000/v_"bottom"-1000-mgsin theta$
$g$ $g$ is acceleration due to gravity and $θ$ $theta$ is the angle of the incline.
$\Rightarrow$ $=>$ Net force upwards $= \frac{30000}{v_{bottom}} - 1000 - 1250 \times 9.81 \times \frac{30}{500}$ $=30000/v_"bottom"-1000-1250xx9.81xx30/500$
Using Newton's Second Law of motion

This must be $= Mass of car \times a_{bottom} = 1250 \times 4$ $="Mass of car" xxa_"bottom"=1250xx4$

Therefore we get

$\frac{30000}{v_{bottom}} - 1000 - 735.75 = 5000$ $30000/v_"bottom"-1000-735.75=5000$
$\Rightarrow \frac{30000}{v_{bottom}} = 6735.75$ $=>30000/v_"bottom"=6735.75$
$=>v_"bottom"=4.45\ ms^-1$ , rounded to two decimal places

Similarly for $v_"top"$

$30000/v_"top"-1000-735.75=1250xx0.2$
$=>v_"top"=15.11\ ms^-1$ , rounded to two decimal places

Gain in KE $=1/2xx1250(v_"top"^2-v_"bottom"^2)$

(ii) Work Done by car’s engine

$=\ "Gain in KE" + "Gain in PE" + "Work Done against resistance"$

$"Gain in KE"$ from part (i)
$"Gain in PE"\ =mgh=1250xx9.81xx30=367875\ J$
$"Work Done against resistance"="resistance"xx"distance"$
$=1000xx500=500000\ J$

(I have used $g=9.81\ ms^-2$ . In some text books show a value of $g=10ms^-2$ ) Choose value of your liking.

Hope this helps.

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It is about finding kinetic energy want help in part 1 not part 2 in part 1 please tell how to find velocity?

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